Equivalence of definitions of Frattini subgroup

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This article gives a proof/explanation of the equivalence of multiple definitions for the term Frattini subgroup
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

Definition in terms of maximal subgroups

For a group G, the Frattini subgroup \Phi(G) is defined as the intersection of all maximal subgroups of G:

\Phi(G) = \bigcap_{M \le_{max} G} M

When there are no maximal subgroups, we define \Phi(G) as the whole group G

Definition as the set of nongenerators

For a group G, the Frattini subgroup \Phi(G) is defined as the set of all nongenerators. An element x \in G is termed a nongenerator if, whenever S \subset G is such that S \cup \{ x \} generates G, S itself generates G.

Facts used


The set of nongenerators is contained in every maximal subgroup

Given: A group G, a nongenerator x for G, a maximal subgroup M of G

To prove: x \in M

Proof: Suppose (contradiction point) x \notin M. Then, \langle M,x \rangle = G. Thus, the set M \cup \{ x \} is a generating set for G. By the definition of nongenerator, we see that M, as a set, is a generating set for G. But this is a contradiction since the subgroup generated by M is M itself.

Thus, x \in M is forced.

Any element in the intersection of all maximal subgroups is a nongenerator

Given: A group G, an element x in the intersection of maximal subgroups of G, a subset S such that S \cup \{ x \} generates G

To prove: S generates G

Proof: Let H be the subgroup generated by S. Suppose (contradiction point) H is proper. Then, x \notin H, since \langle H,x \rangle = G. Thus, H is a 1-completed subgroup: it, along with one outside element, generates the whole group. By the above fact, there exists a maximal subgroup M containing H. But by assumption, x is in every maximal subgroup, so x \in M, so \langle H,x \rangle \le M, so G \le M, a contradiction.

Thus, the assumption that H is proper is false, so H = G, so S generates G.


Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Exercise 25, Page 199 (Section 6.2)